IB HL www.ibmaths.com Adrian Sparrow Factor and remainder theorem.

Slides:

Advertisements

Similar presentations

Aim: How do we solve polynomial equations using factoring?

Advertisements

6.4 Solving Polynomial Equations

Algebra Factorising and cancelling (a 2 – b 2 ) = (a – b)(a + b) (a  b) 2 = a 2  2ab + b 2.

Dividing Polynomials Objectives

Factorising polynomials

Factorising polynomials This PowerPoint presentation demonstrates two methods of factorising a polynomial when you know one factor (perhaps by using the.

The factor theorem The Factor Theorem states that if f(a) = 0 for a polynomial then (x- a) is a factor of the polynomial f(x). Example f(x) = x 2 + x -

Lesson 3- Polynomials 17 May, 2015ML3 MH Objectives : - Definition - Dividing Polynomials Next Lesson - Factor Theorem - Remainder Theorem.

Solving Equations by factoring Algebra I Mrs. Stoltzfus.

Solving quadratic equations Factorisation Type 1: No constant term Solve x 2 – 6x = 0 x (x – 6) = 0 x = 0 or x – 6 = 0 Solutions: x = 0 or x = 6 Graph.

Polynomial Long and Synthetic Division Pre-Calculus.

EXAMPLE 1 Use polynomial long division

Forms of a Quadratic Equation

5.5 Apply the Remainder and Factor Theorem

6.5 – Solving Equations with Quadratic Techniques.

1 Polynomial Functions Exploring Polynomial Functions Exploring Polynomial Functions –Examples Examples Modeling Data with Polynomial Functions Modeling.

Solving Quadratic Equations by Factoring

Additional Mathematics for the OCR syllabus - Algebra 8

Algebra B. Factoring an expression is the opposite of multiplying. a ( b + c ) ab + ac Multiplying Factoring Often: When we multiply an expression we.

Chapter 5: Polynomials & Polynomial Functions

The Quadratic Formula. What does the Quadratic Formula Do ? The Quadratic formula allows you to find the roots of a quadratic equation (if they exist)

Ch 2.5: The Fundamental Theorem of Algebra

2.5 Apply the Remainder and Factor Theorems p. 120 How do you divide polynomials? What is the remainder theorem? What is the difference between synthetic.

EXAMPLE 1 Find a common monomial factor Factor the polynomial completely. a. x 3 + 2x 2 – 15x Factor common monomial. = x(x + 5)(x – 3 ) Factor trinomial.

Factor: Factor: 1. s 2 r 2 – 4s 4 1. s 2 r 2 – 4s b b 3 c + 18b 2 c b b 3 c + 18b 2 c 2 3. xy + 3x – 2y xy + 3x – 2y -

Calculus 3.4 Manipulate real and complex numbers and solve equations AS

Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor The quotient will be here. 2x³ + 3x² - x + 1 is the dividend.

5.3 – Solving Quadratic Equations by Factoring. Ex. 1 Solve y = x 2 + 5x + 6 by factoring.

1 Use the Remainder Theorem and the Factor Theorem. 2.3 Day 2 What You Should Learn.

The Remainder Theorem A-APR 2 Explain how to solve a polynomial by factoring.

Example 1A Solve the equation. Check your answer. (x – 7)(x + 2) = 0

Beginning Algebra 5.7 Solving Equations by Factoring:

Dividing polynomials This PowerPoint presentation demonstrates two different methods of polynomial division. Click here to see algebraic long division.

Warmup Divide using synthetic division using the zero given. Then factor the answer equation completely and solve for the remaining zeroes. Show.

Topic VII: Polynomial Functions 7.3 Solving Polynomial Equations.

Chapter 5 Section 5. EXAMPLE 1 Use polynomial long division Divide f (x) = 3x 4 – 5x 3 + 4x – 6 by x 2 – 3x + 5. SOLUTION Write polynomial division.

Factor Theorem 6 = 2 x 3, we say 2 and 3 are factors of 6, i.e, remainder = 0 Likewise, we say (x+2) and (x – 2) are factors of So, when (x+2) is a factor.

Bellwork 1)Use the quadratic formula to solve 2)Use synthetic substitution to evaluate for x = 2 3) A company’s income is modeled by the function. What.

Remainder and Factor Theorem

ALGEBRA 3 Polynomial Division.

Divide by x - 1 Synthetic Division: a much faster way!

Polynomial and Synthetic Division

7.3 Solving Equations Using Quadratic Techniques

The Quadratic Formula..

Factoring Polynomials

Quadratic Formula Solving for X Solving for quadratic equations.

Factorising polynomials

Remainder Theorem What’s left over?.

Factor Theorem.

Solve a literal equation

PROGRAMME F6 POLYNOMIAL EQUATIONS.

Apply the Remainder and Factor Theorems Lesson 2.5

f(x) = a(x + b)2 + c e.g. use coefficients to

Solving harder linear Simultaneous Equations

Use back - substitution to solve the triangular system. {image}

Know to check all solutions

5.5 Real Zeros of Polynomial Functions

Unit 23 Algebraic Manipulation

Algebra and Functions.

Solving Polynomial Equations

The Quadratic Formula..

A, b and c can be any numbers

The Quadratic Formula..

3.2 The Remainder Theorem.

A, b and c can be any numbers

Polynomials Thursday, 31 October 2019.

A, b and c can be any numbers

Factorise and solve the following:

Presentation transcript:

IB HL Adrian Sparrow Factor and remainder theorem

Factors and solutions Factorise the quadratic x 2 -x-12.(x+3)(x-4) Solve x 2 -x-12=0x=-3 and x=4 Substitute x=-3 into the equation, and calculate the answer. Repeat with x=4. Both answers are 0, because these are factors.

Cubic polynomials This technique will also work for higher order polynomials such as cubics. Show that (x-3) is a factor x 3 -4x 2 +x+6. Substitute in x=3.(3 3 )-4(3 2 )+3+6=0 Hence, using trial and error find the other 2 factors of x 3 -4x 2 +x+6. Hint: these factors are usually integers, start with 0, then try 1, -1, 2, etc. (x-2) (x+1)

Cubic polynomials - typical questions Given that (x-2) and (x+1) are factors of the cubic polynomial 2x 3 +ax 2 +bx+6, find the values of a and b. Substitute in the first factor x=2, make up a linear equation. Substitute in the second factor x=(-1), make up another linear equation. Solve simultaneously to find a and b. 2(2 3 )+(2 2 )a+2b+6=0 2(-1 3 )+(-1 2 )a-b+6=0 16+4a+2b+6=0 4a+2b= a-b+6=0 a-b=-4 a=-5, b=-1

Dividing by factors - quadratics This technique is not used by the IB, but is useful to know and explains the remainders - division by an algebraic factor. Divide the quadratic x 2 -x-12 by one of it’s factors, e.g. (x+3). Divide x 2 by x. Take the answer and multiply by 3. Take this away from the next component in the polynomial. Divide this answer by x. Repeat the steps above. - Two things to note: Remainder = 0 The other factor is given as the answer. Now repeat this process but divide the quadratic by (x-4)

Dividing by factors - cubics Divide the polynomial x 3 -4x 2 +x+6 by (x-3). The quotient left as an answer can now be factorised to give the other two factors of the original cubic. All 3 factors are:

Remainders when dividing Divide the polynomial x 3 -x 2 -10x-8 by (x-2). We have a remainder. This tells us that (x-2) is not a factor of the polynomial. Substitute x=2 into the polynomial x 3 -x 2 -10x-8. Answer = -24

Using the remainder Given that (x+1) is one of the factors of the cubic polynomial ax 3 +5x 2 +bx-9, and that when the polynomial is divided by (x-2) a remainder of 15 is left, find the values of a and b. Substitute in the first factor x=- 1, make up a linear equation. Substitute in the value x=2, but the answer is 15, not 0. Make up a linear equation. Solve simultaneously to find a and b. a(-1 3 )+5(-1 2 )-b-9=0 a(2 3 )+5(2 2 )+2b-9=15 -a+5-b-9=0 a+b=-4 8a+20+2b-9=15 8a+2b=4 a=2, b=-6

Exam questions 2. Given that (x+1) is one of the factors of the cubic polynomial 3x 3 +ax 2 +bx+4, and that when the polynomial is divided by (x-1) a remainder of -12 is left, find the values of a and b. 1. Given that (x-2) and (x-5) are both factors of the cubic polynomial ax 3 +bx 2 -x+30, find the values of a and b. 3. a) Show that (x-3) is one of the factors of the cubic polynomial 2x 3 -3x 2 -14x+15. a=2, b=-11 a=-10, b=-9 b) Hence find the remaining two factors of the cubic polynomial 2x 3 -3x 2 -14x+15. 2(3 3 )-3(3 2 )-14(3)+15=0 Hint: you will need to divide by (x-3). (x-3)(x-1)(2x+5)